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81 lines (71 loc) · 2.5 KB
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class Solution:
def lengthOfLongestSubstring(self, s: str) -> int:
if len(s) < 2:
return len(s)
i = 0
maxlen = 0
dic = {}
for j in range(len(s)):
c = s[j]
if dic.__contains__(c):
j_prev = dic[c]
for k in range(i, j_prev):
del dic[s[k]]
if j - i > maxlen:
maxlen = j - i
i = j_prev + 1
dic[c] = j
return max(maxlen, j - i + 1)
def merge(self, intervals):
if len(intervals) < 2:
return intervals
intervals.sort(key=lambda x:x[0])
itv = intervals[0]
results = []
for i in intervals[1:]:
if i[0] <= itv[1]:
if i[1] > itv[1]:
itv[1] = i[1]
else:
results.append(itv)
itv = i
results.append(itv)
return results
def insert(self, intervals, newInterval):
if not intervals:
return [newInterval]
result = []
for i in range(len(intervals)):
ivt = intervals[i]
if newInterval[1] < ivt[0]:
result.append(newInterval)
result.extend(intervals[i:])
return result
if newInterval[1] <= ivt[1]:
if newInterval[0] < ivt[0]:
ivt[0] = newInterval[0]
result.extend(intervals[i:])
return result
else:
if newInterval[0] <= ivt[0]:
continue
if newInterval[0] <= ivt[1]:
newInterval[0] = ivt[0]
else:
result.append(ivt)
return newInterval.append(newInterval)
def crt_sieving(mods, remainders):
dic = {}
for m, r in zip(mods, remainders):
dic[m] = r # 把模和余数放进字典里,key是模,value是余数
mods.sort(reverse=True) # 把模倒序
adding = mods[0] # 取最大的模,作为每次的加数
num = dic[adding] # 其余数作为初始数字
for mod in mods[1:]: # 依次取剩下的模
re = dic[mod] # 及其余数
while num % mod != re: # num对mod的余数满足条件为止
num += adding # 不满足则一直加上adding
adding *= mod # 满足后就讲adding乘上mod
return num
if __name__ == "__main__":
pass